First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O
n(CaCO₃)=m(CaCO₃)/M(CaCO₃)
n(CaCO₃)=13.00/100.09=0.1299 mol
Δm=13.00+52.65-60.32=5.33 g
m(CO₂)=5.33 g
n(CO₂)=5.33/44.01=0,1211 mol
w=0.1211/0.1299=0,9323 (93.23%)
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce
g of (NH₄)₂S
25g of NH₃ will produce
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
Answer:
203 grams
Explanation:
<em>It is known that 1.0 mole of a compound contains Avogadro's number of molecules (6.022 x 10²³).
</em>
<em><u>Using cross multiplication:</u></em>
1.0 mol contains → 6.022 x 10²³ molecules.
??? mol contains → 7.2 x 10²⁴ molecules.
∴ The no. of moles of (6.3 x 10²⁴ molecules) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.
<em>∴ The no. of grams of NH₃ present = no. of moles x molar mass </em>= (11.96 mol)(17.0 g/mol) = <em>203.3 g ≅ 203.0 g.</em>